Derive an expression for the intensity of the electric field at a point on the equatorial plane of an electric dipole.

Consider an electric dipole consisting of charges - q and + q separated by a distance 2 a . Let E A be the electric field at point P due to charge - q and E B be the electric field at point P due to charge + q where P is a point on the equatorial line of the dipole whose electric field is to be measured. Let r be the distance between the centre of the dipole and point P . Now, resultant electric field at point P would be, E = E A + E B E A = 1 4 π ε 0 q A P 2 ⇒ 1 4 π ε 0 q r 2 + a 2 ( ∵ According to Pythagoras, A P = r 2 + a 2 2 ) E B = 1 4 π ε 0 q B P 2 ⇒ 1 4 π ε 0 q r 2 + a 2 ( ∵ According to Pythagoras, B P = r 2 + a 2 2 ) Now, according to parallelogram law of vector addition, the resultant electric field would become, E = E A 2 + E B 2 + 2 E A E B cos 2 θ Now, since E A = E B we can say, E A = E B = E ' E = E ' 2 + E ' 2 + 2 E ' E ' cos 2 θ ⇒ 2 E ' 2 ( 1 + cos 2 θ ) ⇒ E ' 2 ( 1 + cos 2 θ ) ⇒ E ' 2 2 cos 2 θ E = 2 E ' cos θ - - - ( i ) Now, from △ A O P , cos θ = a r 2 + a 2 Also, we know that, E ' = 1 4 π ε 0 q r 2 + a 2 Putting both these values in equation ( i ) , we get, E = 2 4 π ε 0 q r 2 + a 2 a r 2 + a 2 E = p 4 π ε 0 r 2 + a 2 3 2 Since we know that in dipoles, r > > > 2 a , then a 2 can be neglected. ∴ E = p 4 π ε 0 r 3 2

Derive an expression for the intensity of the electric field at a point on the axial line of an electric dipole.

Consider an electric dipole consisting of charges - q and + q which are separated by a distance of 2 a with a point O being the centre. Let a point P be at the axial line of dipole whose electric field is to be calculated. Let r be the distance between P and O . Let E A be the electric field due to charge - q and E B be the electric field due to charge + q . E A = 1 4 π ε 0 q A P 2 ⇒ 1 4 π ε 0 q a + r 2 E B = 1 4 π ε 0 q B P 2 ⇒ 1 4 π ε 0 q r - a 2 Since E A and E B are opposite to each other and E B > E A , then the resultant intensity at P would be, E = E B - E A ⇒ 1 4 π ε 0 q r - a 2 - 1 4 π ε 0 q r + a 2 ⇒ q 4 π ε 0 1 r - a 2 - 1 r + a 2 ⇒ q 4 π ε 0 r + a 2 - r - a 2 r 2 - a 2 2 ⇒ q 4 π ε 0 4 r a r 2 - a 2 2 ⇒ q 4 π ε 0 2 a 2 r r 2 - a 2 2 ⇒ 1 4 π ε 0 2 p r r 2 - a 2 2 Also, if r > > > a , then, E = 1 4 π ε 0 2 p r 3 .

State Gauss Law of electrostatics and explain it's importance.

Gauss's law states that total electric flux through any closed surface is equal to 1 ∈ 0 times the net charge enclosed by the surface. Total flux ϕ = ∮ E . → d s → = q ∈ 0 Importance : (1) Gauss's law is very useful in calculating the electric field in case of problems where it is possible to construct a closed surface by making a Gaussian surface. (2) Gauss's law is true for any closed surface, no matter what its shape or size. (3) Symmetric considerations in many problems make the application of Gauss's law much easier.

E M B I B E

Intermediate Second Year Physics>Chapter 4 - Electric Charges and Fields>MODEL QUESTIONS (IPE PATTERN)>Q 3

EASY

12th Andhra Pradesh Board

IMPORTANT

Earn 100

How many electrons constitute $1 C$ of charge?

Important Questions on Electric Charges and Fields

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12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Physics>Chapter 4 - Electric Charges and Fields>MODEL QUESTIONS (IPE PATTERN)>Q 4

What happens to the weight of the body when it is charged positively?

EASY

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Physics>Chapter 4 - Electric Charges and Fields>MODEL QUESTIONS (IPE PATTERN)>Q 5

What happens to the force between two charges if the distance between them is :

1) Halved

2)Doubled

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12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Physics>Chapter 4 - Electric Charges and Fields>MODEL QUESTIONS (IPE PATTERN)>Q 6

The electric field lines of force do not intersect.Why?

EASY

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Physics>Chapter 4 - Electric Charges and Fields>MODEL QUESTIONS (IPE PATTERN)>Q 7

Consider two charges

+ q

and

- q

placed at

B

and

C

of an equilateral triangle

A B C

. For this system total charge is zero. But the Electric field (Intensity) at a point

A

which is at equidistant from

B

and

C

is not zero.Why?

HARD

12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Physics>Chapter 4 - Electric Charges and Fields>MODEL QUESTIONS (IPE PATTERN)>Q 8

Electrostatic field line of force do not form closed loops. If they form closed loops then the work done in moving a charge along a closed path will not be zero. From the above two statements, Can you guess the nature of electrostatic force? (conservative/ non-conservative)

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12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Physics>Chapter 4 - Electric Charges and Fields>MODEL QUESTIONS (IPE PATTERN)>Q 9

State Gauss's Law in electrostatics.

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12th Andhra Pradesh Board

IMPORTANT

Intermediate Second Year Physics>Chapter 4 - Electric Charges and Fields>MODEL QUESTIONS (IPE PATTERN)>Q 10

When is the electric flux negative and when is it positive?

EASY

12th Andhra Pradesh Board